Question

There are some CPU chips scrapped by Intel. Each chip is functional with probability p ∈ (0, 1) (and malfunctional with probability 1−p), independently of other chips. Each functional chip meets the Core i3 standard, Core i5 standard and Core i7 with equal probability 1/3, again, independently of other chips. Now, you purchased 5 such chips. Let E1 be the event that you get at most 2 functional chips. Let E2 be the event that among the 5 chips you purchased, there is at least one i3 chip and there is at least one i5 chip and there is at least one i7 chip. (a) Compute P(E1). 2 (b) Compute P(E2). Hint: Inclusion-exclusion. Also, what is the probability that a chip is a functional i3? Also, you may find the set theory formula (A ∪ B ∪ C) c = Ac ∩ Bc ∩ C c useful, where Ac denote the complement of set A. (c) Are E1 and E2 independent? Explain your answer

Answer 1

ANSWER::

a)

Let X is a random variable which is the number of functional chips.

The probability that a chip is functional = p

Total number of chips purchased = 5

We want to find the probability

Hence, possible values of X = 0,1,2

Since X is a Bernoulli random variable hence the probability can be calculated by the expression

so

(answer)

b)

The total number of ways in which we can have the 5 chips with respect to the core standard, which is also our sample space

Now after applying the condition that at least one of each variety should be there, we come at the following 6 cases

i) 3 * i3 , 1 * i5 and  1 * i7

Total number of ways in which we can have 3 i3 chips, 1 i5 chip and 1 i7 chip among the 5 chips

ii) 1 * i3 , 3 * i5 and  1 * i7

Total number of ways in which we can have 1 i3 chip, 3 i5 chip and 1 i7 chip among the 5 chips

iii) 1 * i3 , 1 * i5 and 3 * i7

Total number of ways in which we can have 1 i3 chip, 1 i5 chip and 3 i7 chips among the 5 chips

iv) 2 * i3, 2 * i5 and 1 * i7

Total number of ways in which we can have 2 i3 chip, 2 i5 chip and 1 i7 chip among the 5 chips

v) 1 * i3, 2 * i5 and 2 * i7

Total number of ways in which we can have 1 i3 chip, 2 i5 chips and 2 i7 chips among the 5 chips

vi) 2 * i3, 1 * i5 and 1 * i7

Total number of ways in which we can have 2 i3 chip, 1 i5 chip and 2 i7 chips among the 5 chips

Hence, the total number of ways in which the 5 chips can be categorized following the condition given in the question

= 20+20+20+30+30+30 = 150

So, the required probability of the event E2

answer

The probability that a chip is functional i3 is

= The probability that it is functional * The probability that it is i3

= p * 1/3 = p/3

c)

The probability that a chip is functional is p and this probability is not dependent on the probability 1/3 which is the probability of the chip being i3, i5, or i7. Hence, the event E1 which counts the number of functional chips is an independent event and is not dependent on E2 which counts the number of chips based on their core standard.

The probability for E1 and E2 calculated are independent of each other so the Events E1 and E2 are independent events.

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